5.14 Rigid bar ABCD is loaded and supported as shown in Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.625 in.2 and bar (2) has a cross-sectional area of 1.25 in.2. After load P is applied, the strain in bar (2) is found to be 900 με. Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D. (c) the load P.
Fig. P5.14
Solution From the strain in bar (2), the elongation in bar (2) is e2 = ε 2 L2 = (900 × 10−6 in./in.)(75 in.) = 0.067500 in. Since the joint at C is a perfect connection, the rigid bar deflection at C must equal the elongation of bar (2): vC = e2 = 0.067500 in. ↓
From a deformation diagram of the rigid bar, the vertical deflection of joint B is related to C by similar triangles: v vB = C 36 in. 72 in. 36 in. 1 ∴ vB = vC = vC 72 in. 2 1 = (0.0675 in.) = 0.033750 in. ↓ 2 The joint at B is also a perfect connection; therefore, the downward displacement of B also causes an equal elongation in bar (1): e1 = vB = 0.033750 in. (a) Now that the elongations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is FL σ L e E (0.033750 in.)(30,000 ksi) ∴σ 1 = 1 1 = = 20.25 ksi (T) Ans. e1 = 1 1 = 1 1 50 in. A1 E1 E1 L1 and the normal stress in bar (2) is σ L FL eE (0.067500 in.)(30,000 ksi) ∴σ 2 = 2 2 = = 27.0 ksi (T) Ans. e2 = 2 2 = 2 2 75 in. A2 E2 E2 L2 Alternatively, this stress could be calculated from Hooke’s Law: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
σ 2 = E2ε 2 = (30, 000 ksi)(900 ×10−6 in./in.) = 27.0 ksi (T) (b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C by similar triangles: v vB v = C = D 36 in. 72 in. 96 in. 96 in. 4 4 ∴ vD = vC = vC = (0.0675 in.) = 0.0900 in. ↓ Ans. 72 in. 3 3 (c) The force in each bar can be determined from the stresses computed previously: F1 = σ 1 A1 = (20.25 ksi)(0.625 in.2 ) = 12.65250 kips F2 = σ 2 A2 = (27.0 ksi)(1.25 in.2 ) = 33.75000 kips
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A can be used to determine the load P: ΣM A = (36 in.)F1 + (72 in.)F2 − (96 in.)P = 0 ∴P =
(36 in.)(12.65250 kips) + (72 in.)(33.750 kips) = 30.057 kips = 30.1 kips 96 in.
Ans.
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5.15 Rigid bar ABCD is loaded and supported as shown in Fig. P5.15. Bars (1) and (2) are unstressed before the load P is applied. Bar (1) is made of bronze [E = 100 GPa] and has a crosssectional area of 400 mm2. Bar (2) is made of aluminum [E = 70 GPa] and has a cross-sectional area of 600 mm2. After the load P is applied, the force in bar (1) is found to be 60 kN (in compression). Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A. (c) the load P.
Fig. P5.15
Solution Given that the axial force in bar (1) is 60 kN (in compression), the elongation (i.e., contraction in this case) can be computed as: FL (−60, 000 N)(840 mm) e1 = 1 1 = = −1.260 mm A1 E1 (400 mm 2 )(100, 000 N/mm 2 ) Since the pin at B is a perfect connection, the deflection of the rigid bar at B is equal to the contraction of bar (1): vB = e1 = 1.260 mm ↓ From a deformation diagram of the rigid bar, the vertical deflection of joint B is related to C by similar triangles: v vB = C 3m 1m 1m 1 ∴ vC = vB = vB 3m 3 1 = (1.260 mm) = 0.420 mm ↓ 3 The joint at C is also a perfect connection; therefore, the downward displacement of C also causes an equal elongation in bar (2): e2 = vC = 0.420 mm Note that a downward displacement at C causes elongation (and hence, tension) in bar (2). (a) The normal stress in bronze bar (1) can be computed from the known force in the bar: F −60, 000 N σ1 = 1 = = −150.0 MPa = 150.0 MPa (C) A1 400 mm 2 The normal stress in aluminum bar (2) can be computed from its elongation:
Ans.
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e2 =
F2 L2 σ 2 L2 = A2 E2 E2
∴σ 2 =
e2 E2 (0.420 mm)(70,000 N/mm 2 ) = = 31.956522 MPa = 32.0 MPa (T) L2 920 mm
Ans.
(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A is related to joint B by similar triangles: vA v = B 4m 3m 4m 4 4 ∴ v A = vB = vB = (1.260 mm) = 1.680 mm ↓ Ans. 3m 3 3 (c) The force in bar (1) is given as F1 = −60 kN. The force in bar (2) can be determined from the stress computed previously: F2 = σ 2 A2 = (31.956522 N/mm 2 )(600 mm 2 ) = 19,174 N = 19.174 kN
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about D can be used to determine the load P: ΣM D = (3 m)F1 − (1 m)F2 + (4 m)P = 0 (1 m)(19.174 kN) − (3 m)( − 60 kN) 4m = 49.7935 kN = 49.8 kN
∴P =
Ans.
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5.16 In Fig. P5.16, bronze [E = 100 GPa] links (1) and (2) support rigid beam ABC. Link (1) has a cross-sectional area of 300 mm2 and link (2) has a cross-sectional area of 450 mm2. For an applied load of P = 70 kN, determine the rigid beam deflection at point B.
Fig. P5.16
Solution From a FBD of the rigid beam, write two equilibrium equations: ΣFy = F1 + F2 − 70 kN = 0
(a)
ΣM A = (1, 460 mm)F2 − (540 mm)(70 kN) = 0 (b) Solve Eq. (b) for F2: (540 mm)(70 kN) F2 = = 25.8904 kN 1, 460 mm and backsubstitute into Eq. (a) to obtain F1: F1 = 70 kN − F2 = 70 kN − 25.8904 kN = 44.1096 kN Next, determine the elongations in links (1) and (2): FL (44,109.6 N)(2,000 mm) e1 = 1 1 = = 2.9406 mm A1 E1 (300 mm 2 )(100, 000 N/mm 2 ) FL (25,890.4 N)(3,000 mm) e2 = 2 2 = = 1.7260 mm A2 E2 (450 mm 2 )(100, 000 N/mm 2 ) Since the connections at A and C are perfect, the rigid beam deflections at these joints are equal to the elongations of links (1) and (2), respectively: v A = e1 = 2.9406 mm and vC = e2 = 1.7260 mm
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The deflection at B can be determined from similar triangles: v A − vC v −v = B C 1, 460 mm 920 mm
Solve this expression for vB: 920 mm (v A − vC ) + vC vB = 1,460 mm 920 mm (2.9406 mm − 1.7260 mm) + 1.7260 mm = 1,460 mm = (0.630137)(1.2146 mm) + 1.7260 mm = 2.49 mm ↓
Ans.
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5.17 Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown in Fig P5-17. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of D1 = 0.375 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of D2 = 0.625 in. Aluminum rod (3) has a diameter of D3 = 1.0 in. The yield strength of the bronze is 50 ksi and the yield strength of the aluminum is 36 ksi. (a) Determine the magnitude of load P that can safely be applied to the structure if a minimum factor of safety of 1.5 is required. (b) Determine the deflection of point D for the load determined in part (a). (c) The pin used at B has an ultimate shear strength of 70 ksi. If a factor of safety of 3.0 is required for this double shear pin connection, determine the minimum pin diameter that can be used at B.
Fig. P5.17
Solution Before beginning, the cross-sectional areas of the three rods can be calculated from the specified diameters: A1 = 0.110447 in.2 A2 = 0.306796 in.2 A3 = 0.785398 in.2 The allowable stress of the bronze is σ 50 ksi σ allow,1 = Y ,bronze = = 33.3333 ksi FS 1.5 and the allowable stress of the aluminum is σ 36 ksi σ allow,2 = σ allow,3 = Y ,alum = = 24 ksi FS 1.5 (a) From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P. From a FBD of the rigid bar, write two equilibrium equations: ΣFy = F1 + F2 − F3 = F1 + F2 − P = 0
(a)
ΣM A = (4 ft)F2 − (2.5 ft)F3 = (4 ft)F2 − (2.5 ft)P = 0 (b) Solve Eq. (b) for F2 in terms of the unknown load P: 2.5 ft F2 = P = 0.6250 P (c) 4 ft Backsubstitute this expression into Eq. (a) to obtain F1 in terms of the unknown load P: F1 = P − F2 = P − 0.6250 P = 0.3750 P (d)
Rearrange Eqs. (c) and (d) to express P in terms of F2 and F1, respectively: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P = 1.6000 F2 (e) P = 2.6667 F1 (f) Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod (1) is: F1 ≤ σ allow,1 A1 = (33.3333 ksi)(0.110447 in.2 ) = 3.6816 kips Thus, from Eq. (f), the corresponding maximum load P is: P ≤ 2.6667 F1 = 2.6667(3.6816 kips) = 9.8177 kips
(g)
Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for the aluminum is: F2 ≤ σ allow,2 A2 = (24 ksi)(0.306796 in.2 ) = 7.3631 kips which leads to a corresponding maximum load P from Eq. (e): P ≤ 1.6000 F2 = 1.6000(7.3631 kips) = 11.7810 kips
(h)
Finally, we should also check the capacity of rod (3): F3 ≤ σ allow,3 A3 = (24 ksi)(0.785398 in.2 ) = 18.8496 kips which means that P ≤ 18.8496 kips
(i)
From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is Pmax = 9.8177 kips = 9.82 kips Ans. Based on this maximum load P, the internal forces in the three rods are: F1 = 0.3750 P = 0.3750(9.8177 kips) = 3.6816 kips F2 = 0.6250 P = 0.6250(9.8177 kips) = 6.1361 kips F3 = P = 9.8177 kips (b) Next, determine the elongations in rods (1), (2), and (3): FL (3.6816 kips)(6 ft)(12 in./ft) e1 = 1 1 = = 0.1600 in. A1 E1 (0.110447 in.2 )(15, 000 ksi) FL (6.1361 kips)(8 ft)(12 in./ft) e2 = 2 2 = = 0.1920 in. A2 E2 (0.306796 in.2 )(10, 000 ksi) FL (9.8177 kips)(3 ft)(12 in./ft) e3 = 3 3 = = 0.0450 in. A3 E3 (0.785398 in.2 )(10, 000 ksi) Since the connections at A and C are perfect, the rigid bar deflections at these joints are equal to the elongations of rods (1) and (2), respectively: v A = e1 = 0.1600 in. and vC = e2 = 0.1920 in.
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The rigid bar deflection at B can be determined from similar triangles: vC − v A vB − v A = 4 ft 2.5 ft Solve this expression for vB: 2.5 ft (vC − v A ) + v A vB = 4 ft 2.5 ft (0.1920 in. − 0.1600 in.) + 0.1600 in. = 4 ft = (0.6250)(0.0320 in.) + 0.1600 in. = 0.1800 in. The deflection of joint D is equal to the deflection of the rigid bar at B plus the elongation in rod (3): vD = vB + e3 = 0.1800 in. + 0.0450 in. = 0.2250 in. = 0.225 in. ↓
Ans.
(c) The pin at B has an allowable shear stress of τ 70 ksi τ allow = ult = = 23.3333 ksi FS 3.0 The force tending to shear this pin is equal to the load P = 9.8177 kips. The shear area AV required for this pin is thus V 9.8177 kips AV ≥ = = 0.4208 in.2 τ allow 23.3333 ksi Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional area of the pin: AV = 2 Apin = 2 ×
π
2 Dpin
4 and so the minimum pin diameter is 2×
π
4
2 Dpin ≥ 0.4208 in.2
∴ Dpin ≥ 0.5176 in. = 0.518 in.
Ans.
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5.18 Solve problem 5.14 when there is a clearance of 0.05 in. in the pin connection at C. 5.14 (Repeated here for convenience). Rigid bar ABCD is loaded and supported as shown in Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.625 in.2 and bar (2) has a cross-sectional area of 1.25 in.2. After load P is applied, the strain in bar (2) is found to be 900 με. Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D. (c) the load P.
Fig. P5.14 (repeated)
Solution From the strain in bar (2), the elongation in bar (2) is e2 = ε 2 L2 = (900 × 10−6 in./in.)(75 in.) = 0.067500 in. There is a 0.05-in. clearance in the connection at C; therefore, vC = e2 + 0.05 in. = 0.117500 in. ↓ From a deformation diagram of the rigid bar, the vertical deflection of joint B is related to C by similar triangles: v vB = C 36 in. 72 in. 36 in. 1 ∴ vB = vC = vC 72 in. 2 1 = (0.1175 in.) = 0.05875 in. ↓ 2 The joint at B is a perfect connection; therefore, the downward displacement of B also causes an equal elongation in bar (1): e1 = vB = 0.05875 in. (a) Now that the elongations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is FL σ L e E (0.05875 in.)(30,000 ksi) e1 = 1 1 = 1 1 ∴σ 1 = 1 1 = = 35.25 ksi (T) Ans. A1 E1 E1 L1 50 in. and the normal stress in bar (2) is σ L FL eE (0.067500 in.)(30,000 ksi) e2 = 2 2 = 2 2 ∴σ 2 = 2 2 = = 27.0 ksi (T) Ans. A2 E2 E2 L2 75 in. Alternatively, this stress could be calculated from Hooke’s Law: σ 2 = E2ε 2 = (30, 000 ksi)(900 ×10−6 in./in.) = 27.0 ksi (T) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C by similar triangles: v vB v = C = D 36 in. 72 in. 96 in. 96 in. 4 4 Ans. ∴ vD = vC = vC = (0.1175 in.) = 0.1567 in. ↓ 72 in. 3 3 (c) The force in each bar can be determined from the stresses computed previously: F1 = σ 1 A1 = (35.25 ksi)(0.625 in.2 ) = 22.03125 kips F2 = σ 2 A2 = (27.0 ksi)(1.25 in.2 ) = 33.75000 kips
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A can be used to determine the load P: ΣM A = (36 in.)F1 + (72 in.)F2 − (96 in.)P = 0 ∴P =
(36 in.)(22.03125 kips) + (72 in.)(33.750 kips) = 33.5742 kips = 33.6 kips 96 in.
Ans.
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5.19 The rigid beam in Fig. P5.19 is supported by links (1) and (2), which are made from a polymer material [E = 16 GPa]. Link (1) has a crosssectional area of 400 mm2 and link (2) has a crosssectional area of 800 mm2. Determine the maximum load P that may by applied if the deflection of the rigid beam is not to exceed 20 mm at point C.
Fig. P5.19
Solution Equilibrium: Consider a FBD of the rigid beam and assume tension in each link. ΣFy = − F1 + F2 − P = 0 (a)
ΣM A = (600 mm)F2 − (900 mm)P = 0 From Eq. (b): 900 mm F2 = P = 1.5P 600 mm Backsubstituting into Eq. (a): F1 = F2 − P = 1.5P − P = 0.5P
(b) (c) (d)
Force-deformation relationships: The relationship between internal force and member elongation for links (1) and (2) can be expressed as: FL FL e1 = 1 1 and e2 = 2 2 (e) A1 E1 A2 E2 Geometry of deformations: Consider a deformation diagram of the rigid beam. Using similar triangles, one way to express the relationships between vA, vB, and vC is: v A + vC v −v = C B (f) 900 mm 300 mm Note: Here, vA, vB, and vC are treated as unsigned magnitudes in the directions shown on the deformation diagram.
The rigid beam deflection at A will equal the elongation that occurs in link (1): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
v A = e1 and similarly at B: vB = e2 Equation (f) can now be rewritten in terms of e1 and e2 as: e1 + vC v −e = C 2 900 mm 300 mm or 900 mm (vC − e2 ) = 3vC − 3e2 e1 + vC = 300 mm Solving for vC gives: 2vC = e1 + 3e2 ∴ vC = 0.5e1 + 1.5e2 Substitute the force-deformation relationships from Eq. (e) to obtain: 0.5F1 L1 1.5F2 L2 vC = 0.5e1 + 1.5e2 = + A1 E1 A2 E2 and then substitute Eqs. (c) and (d) to derive an expression for vC in terms of the unknown load P: ⎡ 0.25L1 2.25L2 ⎤ 0.5(0.5P) L1 1.5(1.5P) L2 vC = + = P⎢ + ⎥ A1 E1 A2 E2 A2 E2 ⎦ ⎣ A1 E1 The unknown load P is thus related to the rigid beam deflection at C by: vC P= 0.25L1 2.25L2 + A1 E1 A2 E2 Substituting the appropriate values into this relationship gives the maximum load P that may be applied to the rigid beam at C without causing more than 20 mm deflection: 20 mm P= (0.25)(1,000 mm) (2.25)(1, 250 mm) + 2 2 (400 mm )(16, 000 N/mm ) (800 mm 2 )(16,000 N/mm 2 ) = 77, 283 N = 77.3 kN
Ans.
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5.20 Three aluminum [E = 10,000 ksi] bars are used to support the loads shown in Fig. P5.20. The elongation in each bar must be limited to 0.25 in. Determine the minimum cross-sectional area required for each bar.
Fig. P5.20
Solution Determine the inclination angles for bars (1), (2), and (3): 7 ft ∴θ1 = 34.992° tan θ1 = 10 ft 3 ft ∴θ 2 = 18.435° tan θ 2 = 9 ft 8 ft tan θ3 = ∴θ1 = 57.995° 5 ft The bar lengths are: L1 = (7 ft) 2 + (10 ft) 2 = 12.206556 ft = 146.4787 in. L2 = (3 ft) 2 + (9 ft)2 = 9.486833 ft = 113.8420 in. L3 = (8 ft) 2 + (5 ft) 2 = 9.433981 ft = 113.2078 in.
Equilibrium: Consider a FBD of joint B and write two equilibrium equations: ΣFx = − F1 cos 34.992° + F2 cos18.435° = 0 ΣFy = F1 sin 34.992° + F2 sin18.435° − 31 kips = 0
Solve these two equations simultaneously to obtain F1 = 36.61967 kips and F2 = 31.62278 kips. Next, consider a FBD of joint C and write two additional equilibrium equations: ΣFx = − F2 cos18.435° + F3 cos 57.995° = 0 ΣFy = − F2 sin18.435° + F3 sin 57.995° − 38 kips = 0
From this, the internal force in bar (3) is F3 = 56.60389 kips. The minimum cross-sectional area required for each bar is: F L (36.61967 kips)(146.4787 in.) A1 ≥ 1 1 = = 2.15 in.2 e1 E1 (0.25 in.)(10,000 ksi)
Ans.
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F2 L2 (31.62278 kips)(113.8420 in.) = = 1.440 in.2 e2 E2 (0.25 in.)(10,000 ksi) F L (56.60389 kips)(113.2078 in.) A3 ≥ 3 3 = = 2.56 in.2 e3 E3 (0.25 in.)(10,000 ksi) A2 ≥
Ans. Ans.
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5.21 A tie rod (1) and a pipe strut (2) are used to support an 80-kip load as shown in Fig. P5.21. Pipe strut (2) has an outside diameter of 6.625 in. and a wall thickness of 0.280 in. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of E = 29,000 ksi and a yield strength of σY = 36 ksi. For the tie rod, the minimum factor of safety with respect to yield is 1.5 and the maximum allowable axial elongation is 0.20 in. (a) Determine the minimum diameter D required to satisfy both constraints for tie rod (1). (b) Draw a deformation diagram showing the final position of joint B.
Fig. P5.21
Solution The angles of inclination for members (1) and (2) are: 12 ft tan θ1 = ∴θ1 = 26.565° 24 ft 30 ft tan θ 2 = ∴θ 2 = 51.340° 24 ft The member lengths are: L1 = (12 ft) 2 + (24 ft) 2 = 26.8328 ft = 321.9938 in. L2 = (30 ft)2 + (24 ft) 2 = 38.4187 ft = 461.0249 in.
(a) Equilibrium: Consider a FBD of joint B and write two equilibrium equations: ΣFx = − F1 cos 26.565° − F2 cos 51.340° = 0 ΣFy = F1 sin 26.565° − F2 sin 51.340° − 80 kips = 0
Solve these two equations simultaneously to obtain F1 = 51.1103 kips and F2 = −73.1786 kips. The allowable normal stress for tie rod (1) is σ 36 ksi = 24 ksi σ allow = Y = FS 1.5 The minimum cross-sectional area required to satisfy the normal stress requirement is F 51.1103 kips = 2.1296 in.2 A≥ 1 = σ allow 24 ksi The maximum allowable axial elongation for the tie rod is 0.20 in. The minimum cross-sectional area required to satisfy the deformation requirement is F L (51.1103 kips)(321.9938 in.) A≥ 1 1 = = 2.8374 in.2 e1 E1 (0.20 in.)(29,000 ksi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
To satisfy both the stress and deformation requirements, the tie rod must have a minimum crosssectional area of A = 2.8374 in.2. The corresponding rod diameter is
π
4
2 Drod ≥ 2.8374 in.2
Drod ≥ 1.901 in.
Ans.
(b) The pipe has a cross-sectional area of A2 = 5.5814 in.2. The pipe elongation is FL (−73.1786 kips)(461.0249 in.) e2 = 2 2 = = −0.2084 in. A2 E2 (5.5814 in.2 )(29, 000 ksi) Rod (1) elongates 0.20 in. and pipe (2) contracts 0.2084 in. Therefore, the deformation diagram showing the final position of joint B is shown.
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5.22 Two axial members are used to support a load of P = 72 kips as shown in Fig. P5.22. Member (1) is 12-ft long, it has a cross-sectional area of A1 = 1.75 in.2, and it is made of structural steel [E = 29,000 ksi]. Member (2) is 16-ft long, it has a crosssectional area of A2 = 4.50 in.2, and it is made of an aluminum alloy [E = 10,000 ksi]. (a) Compute the normal stress in each axial member. (b) Compute the elongation of each axial member. (c) Draw a deformation diagram showing the final position of joint B. (d) Compute the horizontal and vertical displacements of joint B.
Fig. P5.22
Solution (a) Consider a FBD of joint B and write two equilibrium equations: ΣFx = − F1 + F2 cos 55° = 0 ΣFy = F2 sin 55° − 72 kips = 0
Solve these two equations simultaneously to obtain F1 = 50.4149 kips and F2 = 87.8958 kips. The normal stress in member (1) is: F 50.4149 kips σ1 = 1 = = 28.8 ksi A1 1.75 in.2 and the normal stress in member (2) is: F 87.8958 kips σ2 = 2 = = 19.53 ksi A2 4.50 in.2 (b) The member elongations are F L (50.4149 kips)(12 ft)(12 in./ft) e1 = 1 1 = = 0.1430 in. A1 E1 (1.75 in.2 )(29, 000 ksi) FL (87.8958 kips)(16 ft)(12 in./ft) e2 = 2 2 = = 0.375 in. A2 E2 (4.50 in.2 )(10, 000 ksi)
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(c) Member (1) elongates 0.1430 in. and member (2) elongates 0.375 in. Therefore, the deformation diagram showing the final position of joint B is shown.
(d) The horizontal displacement of B is Δx = 0.1430 in. The vertical displacement is found from 0.375 in. + 0.1430 cos 55° sin 55° = Δy 0.375 in. + 0.1430 cos 55° 0.375 in. + 0.0820 in. ∴Δy = = = 0.558 in. ↓ sin 55° sin 55°
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Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
